# Thread: Phase 10 Starting hands

1. ## Phase 10 Starting hands

Hello, I just would like someone to work out a number for me and tell me if they get the same answer as me please. I'm getting a very big answer and just want to validate my methodology before continuing with the rest of my analysis.

Problem:

There is a game called Phase 10, which has 108 cards in the deck. Each player has a starting hand of 10 cards.

Given that I only know my own hand and no others, irrelevant of number of players. There are 140,517,366,424,763,000,000 possible combinations of starting hands available to me?

Is this correct? I would like to say I am sure I am but I just want to be double check.

Thank you.

2. ## Re: Phase 10 Starting hands

Originally Posted by RedundantDummy
Problem:
There is a game called Phase 10, which has 108 cards in the deck. Each player has a starting hand of 10 cards.
Given that I only know my own hand and no others, irrelevant of number of players. There are 140,517,366,424,763,000,000 possible combinations of starting hands available to me?.
No it is not if we use the common understanding of cards. You say that the deck contains 108 distinct cards. Your hand consists of the of those cards. So $\dbinom{108}{10}=38722819230810$ SEE HERE.
You see that a hand of cards is determined only by content and not order.

3. ## Re: Phase 10 Starting hands

Unless the cards are 2 regular decks of playing
cards, including the jokers...

4. ## Re: Phase 10 Starting hands

Thank you.

I completely forgot about order, I presume the differences are;

108 x 107 x 106 x 105 x 104 x 103 x 102 x 101 x 100 x 100 = possible combinations including order

where as;

dbinom{108}{10} = possible combinations ignoring order?

5. ## Re: Phase 10 Starting hands

Originally Posted by RedundantDummy
Thank you.

I completely forgot about order, I presume the differences are;

108 x 107 x 106 x 105 x 104 x 103 x 102 x 101 x 100 x 100 = possible combinations including order

where as;

dbinom{108}{10} = possible combinations ignoring order?
if you pick $k$ cards from a deck of $n$ there are

$\dbinom{n}{k}$ unordered combinations and

$k! \dbinom{n}{k} = \dfrac{n!}{(n-k)!}$ ordered combinations